Chapter 12: Problem 42
If \(f(x, y, z)=(x y / z)^{1 / 2}\), find \(f_{x}(-2,-1,8)\).
Short Answer
Expert verified
The partial derivative \( f_x(-2, -1, 8) \) is \( \frac{-1}{8} \).
Step by step solution
01
Understand the Exercise
We are given the function \( f(x, y, z) = \left( \frac{x y}{z} \right)^{1/2} \) and asked to find the partial derivative of \( f \) with respect to \( x \) at the point \((-2, -1, 8)\).
02
Setup Partial Derivative
The partial derivative of \( f(x, y, z) \) with respect to \( x \) is given by \( f_x(x, y, z) = \frac{\partial}{\partial x}\left( \frac{x y}{z} \right)^{1/2} \).
03
Apply Chain Rule
To differentiate \( \left( \frac{x y}{z} \right)^{1/2} \) with respect to \( x \), we'll use the chain rule. Let \( u = \frac{x y}{z} \), thus \( f(x, y, z) = u^{1/2} \). The derivative \( \frac{d}{dx} u^{1/2} = \frac{1}{2}u^{-1/2} \cdot \frac{du}{dx} \).
04
Differentiate the Inner Function
The derivative of \( u = \frac{x y}{z} \) with respect to \( x \) is \( \frac{y}{z} \), since \( y \) and \( z \) are treated as constants in partial differentiation with respect to \( x \).
05
Calculate Derivative at the Point
Substitute \( x = -2 \), \( y = -1 \), and \( z = 8 \) into the formula. First, calculate \( u = \frac{x y}{z} = \frac{-2 \cdot (-1)}{8} = \frac{1}{4} \). Then substitute into the derivative, \( f_x(x, y, z) = \frac{1}{2}(\frac{1}{4})^{-1/2} \cdot \frac{-1}{8} = \frac{1}{2} \cdot 2 \cdot \frac{-1}{8} = \frac{-1}{8} \).
06
Final Answer
The partial derivative \( f_x(-2, -1, 8) \) is \( \frac{-1}{8} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept in multivariable calculus, particularly for functions of several variables like \( f(x, y, z) = (\frac{xy}{z})^{1/2} \). In essence, a partial derivative measures the rate at which a function changes as one of its variables is changed, while keeping other variables constant.
Taking the partial derivative with respect to \( x \) of our given function means evaluating how the function changes as \( x \) changes, treating \( y \) and \( z \) as constants. We find \( f_x(x, y, z) \), which is expressed as \( \frac{\partial}{\partial x}(\frac{xy}{z})^{1/2} \).
The process involves the following steps:
Taking the partial derivative with respect to \( x \) of our given function means evaluating how the function changes as \( x \) changes, treating \( y \) and \( z \) as constants. We find \( f_x(x, y, z) \), which is expressed as \( \frac{\partial}{\partial x}(\frac{xy}{z})^{1/2} \).
The process involves the following steps:
- First, identify the expression inside the square root: \( \frac{xy}{z} \).
- Apply the chain rule to differentiate with respect to \( x \), as detailed next.
Chain Rule
In the context of partial differentiation, the chain rule is crucial when dealing with composite functions. Our function \((\frac{xy}{z})^{1/2}\) can be seen as a composition of the inner function \( u = \frac{xy}{z} \) and the outer function \( u^{1/2} \).
To differentiate this composite function concerning \( x \), we apply the chain rule:
To differentiate this composite function concerning \( x \), we apply the chain rule:
- Differentiate the outer function: For \( u^{1/2} \), this gives \( \frac{1}{2}u^{-1/2} \).
- Then, differentiate the inner function \( u = \frac{xy}{z} \) with respect to \( x \). Since \( y \) and \( z \) are constants, the derivative is \( \frac{y}{z} \).
Functions of Several Variables
Functions of several variables, such as \( f(x, y, z) = (\frac{xy}{z})^{1/2} \), are integral to modeling real-world scenarios involving multiple influencing factors. Each variable represents a dimension, allowing the function to describe a surface or shape in a multidimensional space.
In practical applications:
In practical applications:
- Variables often correspond to different physical quantities (e.g., time, distance, temperature).
- The function's value reflects a property influenced by these quantities, providing insights into how changes in each variable affect the overall outcome.